Question

The value of $3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$ is

Answer 1

Given expression:
$3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)$
Now,
$3(\sin x-\cos x{)}^4=(1-\sin 2x{)}^2=3(1-2\sin 2x+{\sin }^{2}2x)\Rightarrow 3(\sin x-\cos x{)}^4=3(1-2\sin 2x+{\sin }^{2}2x)\cdots (1\left)6\right(\sin x+\cos x{)}^2=6(1+\sin 2x)\cdots \left(2\right)4({\sin }^{6}x+{\cos }^{6}x)=4\left[\right({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)]=4[1-\frac{3}{4}{\sin }^{2}2x]=4-3{\sin }^{2}2x\Rightarrow 4({\sin }^{6}x+{\cos }^{6}x)=4-3{\sin }^{2}2x\cdots (3)$

Using equation $\left(1\right),\left(2\right)$ and $\left(3\right)$,
$3(\sin x-\cos x{)}^4+6(\sin x+\cos x{)}^2+4({\sin }^{6}x+{\cos }^{6}x)=13$